Find derivative by first principle f(x) = xsinx [with Video] - Teachoo

Example 20 (ii) - Chapter 13 Class 11 Limits and Derivatives - Part 2

Example 20 (ii) - Chapter 13 Class 11 Limits and Derivatives - Part 3

  1. Chapter 13 Class 11 Limits and Derivatives
  2. Serial order wise

Transcript

Example 20 Find the derivative of f(x) from the first principle, where f(x) is (ii) x sin x Given f (x) = x sin x We need to find Derivative of f(x) We know that f’(x) = lim┬(hβ†’0) 𝑓⁑〖(π‘₯ + β„Ž) βˆ’ 𝑓(π‘₯)γ€—/β„Ž Here, f (x) = x sin x So, f (x + h) = (x + h) sin (x + h) Putting values f’(x) =lim┬(hβ†’0) ((π‘₯ + β„Ž) sin⁑〖 (π‘₯ + β„Ž) βˆ’ π‘₯ sin⁑〖π‘₯ γ€— γ€—)/β„Ž Using sin (A + B) = sin A cos B + cos A sin B = lim┬(hβ†’0)⁑〖((π‘₯ + β„Ž)(sin⁑〖π‘₯ cosβ‘γ€–β„Ž +γ€– cos π‘₯〗⁑sinβ‘γ€–β„Ž γ€— )γ€— βˆ’ π‘₯ sin⁑π‘₯ γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑〖(π‘₯(sin⁑〖π‘₯ cosβ‘γ€–β„Ž +γ€– cos〗⁑〖π‘₯ sinβ‘γ€–β„Ž) + β„Ž (sin⁑〖π‘₯ cosβ‘γ€–β„Ž +γ€– cos〗⁑〖π‘₯ sinβ‘γ€–β„Ž) βˆ’ π‘₯ sin⁑π‘₯ γ€— γ€— γ€— γ€— γ€— γ€— γ€— γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑〖(π‘₯ sin⁑〖π‘₯ cosβ‘γ€–β„Ž + π‘₯ cos⁑〖π‘₯ sinβ‘γ€–β„Ž + β„Ž γ€–(sin〗⁑〖π‘₯ cosβ‘γ€–β„Ž + cos⁑〖π‘₯ sinβ‘γ€–β„Ž) βˆ’ π‘₯ sin⁑π‘₯ γ€— γ€— γ€— γ€— γ€— γ€— γ€— γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑〖(π‘₯𝑠𝑖𝑛 π‘₯ cosβ‘γ€–β„Ž βˆ’ π‘₯ sin⁑〖π‘₯ + π‘₯ cos⁑〖π‘₯ sinβ‘γ€–β„Ž + β„Ž(sin⁑〖π‘₯ cosβ‘γ€–β„Ž + cos⁑〖π‘₯ sinβ‘β„Ž γ€— γ€—)γ€— γ€— γ€— γ€— γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑〖(π‘₯𝑠𝑖𝑛 π‘₯ γ€–(cosγ€—β‘γ€–β„Ž βˆ’ 1)+ π‘₯ cos⁑〖π‘₯ sinβ‘γ€–β„Ž + β„Ž(sin⁑〖π‘₯ cosβ‘γ€–β„Ž + cos⁑〖π‘₯ sinβ‘β„Ž γ€— γ€—)γ€— γ€— γ€— γ€—)/β„Žγ€— = lim┬(hβ†’0)⁑((π‘₯ sin⁑〖π‘₯ (cosβ‘γ€–β„Ž βˆ’1)γ€— γ€—)/β„Ž+(π‘₯ cos⁑〖π‘₯ sinβ‘β„Ž γ€—)/β„Ž+(β„Ž (sin⁑〖π‘₯ cosβ‘γ€–β„Ž +cos⁑〖π‘₯ sinβ‘γ€–β„Ž)γ€— γ€— γ€— γ€—)/β„Ž) = lim┬(hβ†’0)⁑〖〖x sin〗⁑〖π‘₯ (cosβ‘γ€–β„Ž βˆ’1)γ€— γ€—/β„Ž+lim┬(hβ†’0) (π‘₯ cos⁑〖π‘₯ sinβ‘β„Ž γ€— )/h+lim┬(hβ†’0) (sin⁑〖π‘₯ cosβ‘γ€–β„Ž +cos⁑〖π‘₯ sinβ‘γ€–β„Ž )γ€— γ€— γ€— γ€— γ€— = – x sin x (π₯𝐒𝐦)┬(π‘β†’πŸŽ) ((πŸβˆ’γ€– 𝒄𝒐𝒔〗⁑〖𝒉)γ€—)/𝒉+π‘₯ cos⁑〖π‘₯ (π₯𝐒𝐦)┬(π‘β†’πŸŽ) π’”π’Šπ’β‘π’‰/𝐑+γ€— lim┬(hβ†’0) (sin⁑〖π‘₯ cosβ‘γ€–β„Ž +cos⁑〖π‘₯ sinβ‘γ€–β„Ž )γ€— γ€— γ€— γ€— = – x sin x (0) + x cos x (1) + ( sin x cos 0 + cos x sin 0) = 0 + x cos x + sin Γ— 1 + cos x Γ— 0 = 0 + x cos x + sin x + 0 = x cos x + sin x Hence f’ (x) = x cos x + sin x

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.